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The temperature was 
4^(@)F at 10:00 PM. It dropped an average of 
3^(@) each hour for the next 6 hours. What was the temperature at 4:00 AM?

33. The temperature was 4F 4^{\circ} \mathrm{F} at 1010:0000 PM. It dropped an average of 3 3^{\circ} each hour for the next 66 hours. What was the temperature at 44:0000 AM?

Full solution

Q. 33. The temperature was 4F 4^{\circ} \mathrm{F} at 1010:0000 PM. It dropped an average of 3 3^{\circ} each hour for the next 66 hours. What was the temperature at 44:0000 AM?
  1. Initial Temperature: Initial temperature at 1010:0000 PM: 4°F4\degree F\newlineTemperature drop per hour: 3°F3\degree F\newlineTotal hours of temperature drop: 66 hours\newlineCalculate total temperature drop: 3°F/hour×63\degree F/hour \times 6 hours
  2. Total Temperature Drop: Total temperature drop: 3F/hour×6hours=18F3\,^\circ\mathrm{F}/\text{hour} \times 6\,\text{hours} = 18\,^\circ\mathrm{F}\newlineSubtract total drop from initial temperature: 4F18F4\,^\circ\mathrm{F} - 18\,^\circ\mathrm{F}
  3. Final Temperature: Final temperature at 44:0000 AM: 4°F18°F=14°F4\degree F - 18\degree F = -14\degree F

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